NYU Calculus Class Uses Tangent Lines to Approximate Sine Function

Section 2.8
Linear Approximation and Differentials

V63.0121.002.2010Su, Calculus I

New York University

May 26, 2010

Announcements

Quiz 2 Thursday on Sections 1.5–2.5
No class Monday, May 31
Assignment 2 due Tuesday, June 1

. . . . . .

Announcements

Quiz 2 Thursday on
Sections 1.5–2.5
No class Monday, May 31
Assignment 2 due
Tuesday, June 1

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 2 / 27

Objectives
Use tangent lines to make
linear approximations to a
function.
Given a function and a
point in the domain,
compute the
linearization of the
function at that point.
Use linearization to
approximate values of
functions
Given a function, compute
the differential of that
function
Use the differential
notation to estimate error
in linear approximations. . . . . . .


Outline

The linear approximation of a function near a point
Examples
Questions

Differentials
Using differentials to estimate error

Advanced Examples

. . . . . .


The Big Idea

Question
Let f be differentiable at a. What linear function best approximates f
near a?

. . . . . .


The Big Idea

Question
near a?

Answer
The tangent line, of course!

. . . . . .


The Big Idea

Question
near a?

Answer

Question
What is the equation for the line tangent to y = f(x) at (a, f(a))?

. . . . . .


The Big Idea

Question
near a?

Answer

Question
What is the equation for the line tangent to y = f(x) at (a, f(a))?

Answer

L(x) = f(a) + f′ (a)(x − a)

. . . . . .


The tangent line is a linear approximation

y
.

L(x) = f(a) + f′ (a)(x − a)

is a decent approximation to f L
. (x) .
near a. f
.(x) .

f
.(a) .
.
x−a

. x
.
a
. x
.

. . . . . .


The tangent line is a linear approximation

y
.

L(x) = f(a) + f′ (a)(x − a)

is a decent approximation to f L
. (x) .
near a. f
.(x) .
How decent? The closer x is to
a, the better the approxmation f
.(a) .
.
x−a
L(x) is to f(x)

. x
.
a
. x
.

. . . . . .


Example
.
Example
Estimate sin(61◦ ) = sin(61π/180) by using a linear approximation
(i) about a = 0 (ii) about a = 60◦ = π/3.

.

. . . . . .


Example
.
Example

Solution (i)

If f(x) = sin x, then f(0) = 0
and f′ (0) = 1.
So the linear approximation
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

Solution (i) Solution (ii)
(π)
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3
and f′ (0) = 1. f′ π = .
3
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = .
3
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2
So L(x) =
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So the linear approximation 3 1( π)
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈
180 180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180

Calculator check: sin(61◦ ) ≈
.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180

Calculator check: sin(61◦ ) ≈ 0.87462.
.

. . . . . .


Illustration

y
.

y
. = sin x

. x
.
. 1◦
6
. . . . . .


Illustration

y
.
y
. = L1 (x) = x

y
. = sin x

. x
.
0
. . 1◦
6
. . . . . .


Illustration

y
.
y
. = L1 (x) = x

b
. ig difference! y
. = sin x

. x
.
0
. . 1◦
6
. . . . . .


Illustration

y
.
y
. = L1 (x) = x

√ ( )
y
. = L2 (x) = 2
3
+ 1
2 x− π
3
y
. = sin x
.

. . x
.
0
. .
π/3 . 1◦
6
. . . . . .


Illustration

y
.
y
. = L1 (x) = x

√ ( )
y
. = L2 (x) = 2
3
+ 1
2 x− π
3
y
. = sin x
. . ery little difference!
v

. . x
.
0
. .
π/3 . 1◦
6
. . . . . .


Another Example

Example
√
Estimate 10 using the fact that 10 = 9 + 1.

. . . . . .


Another Example

Example
√

Solution
√
The key step is to use a linear approximation to f(x) =
√ x near a = 9
to estimate f(10) = 10.

. . . . . .


Another Example

Example
√

Solution
√
√ x near a = 9
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1) = ≈ 3.167
2·3 6

. . . . . .


Another Example

Example
√

Solution
√
√ x near a = 9
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1) = ≈ 3.167
2·3 6

( )2
19
Check: =
6
. . . . . .


Another Example

Example
√

Solution
√
√ x near a = 9
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1) = ≈ 3.167
2·3 6

( )2
19 361
Check: = .
6 36
. . . . . .


Dividing without dividing?
Example
Suppose I have an irrational fear of division and need to estimate
577 ÷ 408. I write
577 1 1 1
= 1 + 169 = 1 + 169 × × .
408 408 4 102
1
But still I have to find .
102

. . . . . .


Dividing without dividing?
Example
Suppose I have an irrational fear of division and need to estimate
577 ÷ 408. I write
577 1 1 1
= 1 + 169 = 1 + 169 × × .
408 408 4 102
1
But still I have to find .
102

Solution
1
Let f(x) = . We know f(100) and we want to estimate f(102).
x
1 1
f(102) ≈ f(100) + f′ (100)(2) = − (2) = 0.0098
100 1002
577
=⇒ ≈ 1.41405
408 . . . . . .


Questions

Example
Suppose we are traveling in a car and at noon our speed is 50 mi/hr.
How far will we have traveled by 2:00pm? by 3:00pm? By midnight?

. . . . . .


Answers

Example

. . . . . .


Answers

Example

Answer

100 mi
150 mi
600 mi (?) (Is it reasonable to assume 12 hours at the same
speed?)

. . . . . .


Questions

Example

Example
Suppose our factory makes MP3 players and the marginal cost is
currently $50/lot. How much will it cost to make 2 more lots? 3 more
lots? 12 more lots?

. . . . . .


Answers

Example
lots? 12 more lots?

Answer

$100
$150
$600 (?)

. . . . . .


Questions

Example

Example
lots? 12 more lots?

Example
Suppose a line goes through the point (x0 , y0 ) and has slope m. If the
point is moved horizontally by dx, while staying on the line, what is the
corresponding vertical movement?

. . . . . .


Answers

Example

. . . . . .


Answers

Example

Answer
The slope of the line is
rise
m=
run
We are given a “run” of dx, so the corresponding “rise” is m dx.

. . . . . .


Outline

Examples
Questions

Differentials

Advanced Examples

. . . . . .


Differentials are another way to express derivatives

f(x + ∆x) − f(x) ≈ f′ (x) ∆x y
.
∆y dy

Rename ∆x = dx, so we can
write this as
.
∆y ≈ dy = f′ (x)dx. .
dy
.
∆y

And this looks a lot like the .
.
dx = ∆x
Leibniz-Newton identity

dy .
= f′ (x) x
.
dx x x
. . + ∆x

. . . . . .


Differentials are another way to express derivatives

f(x + ∆x) − f(x) ≈ f′ (x) ∆x y
.
∆y dy

Rename ∆x = dx, so we can
write this as
.
∆y ≈ dy = f′ (x)dx. .
dy
.
∆y

And this looks a lot like the .
.
dx = ∆x
Leibniz-Newton identity

dy .
= f′ (x) x
.
dx x x
. . + ∆x
Linear approximation means ∆y ≈ dy = f′ (x0 ) dx near x0 .
. . . . . .



y
.
If y = f(x), x0 and ∆x is known,
and an estimate of ∆y is
desired:
Approximate: ∆y ≈ dy .
Differentiate: dy = f′ (x) dx .
∆y
.
dy

Evaluate at x = x0 and .
.
dx = ∆x
dx = ∆x.

. x
.
x x
. . + ∆x

. . . . . .


Example
A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting
machine will cut a rectangle whose width is exactly half its length, but
the length is prone to errors. If the length is off by 1 in, how bad can the
area of the sheet be off by?

. . . . . .


Example

Solution
1 2
Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in.
2

. . . . . .


Example

Solution
1 2
2 ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288

. . . . . .


Example

Solution
1 2
2 ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288
dA
(II) = ℓ, so dA = ℓ dℓ, which should be a good estimate for ∆ℓ.
dℓ
When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667. So we
1 8
3
get estimates close to the hundredth of a square foot.

. . . . . .


Why?

Why use linear approximations dy when the actual difference ∆y is
known?
Linear approximation is quick and reliable. Finding ∆y exactly
depends on the function.
These examples are overly simple. See the “Advanced Examples”
later.
In real life, sometimes only f(a) and f′ (a) are known, and not the
general f(x).

. . . . . .


Outline

Examples
Questions

Differentials

Advanced Examples

. . . . . .


Gravitation
Pencils down!

Example

Drop a 1 kg ball off the roof of the Silver Center (50m high). We
usually say that a falling object feels a force F = −mg from gravity.

. . . . . .


Gravitation
Pencils down!

Example

Drop a 1 kg ball off the roof of the Silver Center (50m high). We
usually say that a falling object feels a force F = −mg from gravity.
In fact, the force felt is
GMm
F(r) = − ,
r2
where M is the mass of the earth and r is the distance from the
center of the earth to the object. G is a constant.
GMm
At r = re the force really is F(re ) = = −mg.
r2
e
What is the maximum error in replacing the actual force felt at the
top of the building F(re + ∆r) by the force felt at ground level
F(re )? The relative error? The percentage error? . . . . . .


Gravitation Solution
Solution
We wonder if ∆F = F(re + ∆r) − F(re ) is small.
Using a linear approximation,

dF GMm
∆F ≈ dF = dr = 2 3 dr
dr re re
( )
GMm dr ∆r
= 2
= 2mg
re re re

∆F ∆r
The relative error is ≈ −2
F re
re = 6378.1 km. If ∆r = 50 m,
∆F ∆r 50
≈ −2 = −2 = −1.56 × 10−5 = −0.00156%
F re 6378100
. . . . . .


Systematic linear approximation

√ √
2 is irrational, but 9/4 is rational and 9/4 is close to 2.

. . . . . .



√ √
2 is irrational, but 9/4 is rational and 9/4 is close to 2. So
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12

. . . . . .



√ √
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12

This is a better approximation since (17/12)2 = 289/144

. . . . . .



√ √
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12

This is a better approximation since (17/12)2 = 289/144
Do it again!
√ √ √ 1
2 = 289/144 − 1/144 ≈ 289/144 + (−1/144) = 577/408
2(17/12)
( )2
577 332, 929 1
Now = which is away from 2.
408 166, 464 166, 464

. . . . . .


Illustration of the previous example

.

. . . . . .



.
2
.

. . . . . .



.

.
2
.

. . . . . .



. 2, 17 )
( 12
. .

.
2
.

. . . . . .



.
. 2, 17/12)
(
. . 4, 3)
(9 2

. . . . . .



.
. 2, 17/12)
(
.. ( . 9, 3)
(
)4 2
289 17
. 144 , 12

. . . . . .



.
. 2, 17/12)
(
.. ( . 9, 3)
(
( 577 ) )4 2
. 2, 408 289 17
. 144 , 12

. . . . . .


Summary

Linear approximation: If f is differentiable at a, the best linear
approximation to f near a is given by

Lf,a (x) = f(a) + f′ (a)(x − a)

Differentials: If f is differentiable at x, a good approximation to
∆y = f(x + ∆x) − f(x) is

dy dy
∆y ≈ dy = · dx = · ∆x
dx dx
Don’t buy plywood from me.

. . . . . .


NYU Calculus Class Uses Tangent Lines to Approximate Sine Function

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